Question

For each relation below, determine the following.(i) Is it a function? If not, explain why not and stop. Otherwise, answer part (ii).(ii) What are its domain and image?

(a){(x, y) :x, y∈Z, y- 2x}.

(b){(x, y) :x, y∈Z, xy- 0}.

(c){(x, y) :x, y∈Z, y-x2}.

(d){(x, y) :x, y∈Z, x|y}.

(e){(x, y) :x, y∈Z, x+y= 0}.

(f){(x, y) :x, y∈R, x2+y2= 1}.

Answer 1

(a) {(x,y) : x, y in Z, y = 2x}

It is a function.

The domain is, Z

The image is, 2Z = { 2k : k is in Z } = set of all even integers.

(b) {(x,y) : x, y in Z, xy = 0} = {(x,y) : x, y in Z, either x = 0 or y = 0} = (X-axis) U (Y-axis)

It is not a function, since, for example, if x = 0, we can have different values of y, say, if x = 0, y can be any value in Z. So, it is a relation but not a function.

(c) {(x,y) : x, y in Z, y = x² }

It is a function.

The domain is, Z

The image is the set of all square integers, i.e. { 0, 1, 4, 9, 16, 25, 36, 49, 64, ... } = { k² : k is in Z }

(d) {(x,y) : x, y in Z, x | y }

It is not a function, since, (2,4) & (2,6) belongs to the set among others.

So, the integer has at least two images, which can't happen in a function.

It is a relation, however.

(e) {(x,y) : x, y in Z, x+y = 0, i.e. y = - x }

It is a function.

The domain is, Z

The image is Z

(f) {(x,y) : x, y in Z , x² + y² = 1 }

It is not a function, since, for, x = 0, we have, y = +1 as well as y = - 1

So (0,1) & (0,-1) both are in the set.

So, it is not a function.